Question: Let $R$ be the region enclosed by the line $y=1$, the line $x=4$ and the curve $y=\sqrt{x}+1$. $y$ $x$ ${y=\sqrt{x}+1}$ $ R$ $ 0$ $ 4$ $ 1$ A solid is generated by rotating $R$ about the line $y=1$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Explanation: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\sqrt{x}+1}$ Each slice is a cylinder. Let the width of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\sqrt{x}+1}$ $ 0$ $ 4$ $ 1$ $r$ The radius is equal to the distance between the curve $y=\sqrt{x}+1$ and the line $y=1$. In other words, for any $x$ -value, this is the equation for $r(x)$ : $\begin{aligned} r(x)}&=(\sqrt{x}+1)-(1) \\\\ &=\sqrt{x}} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(\sqrt{x}}\right)^2 \\\\ &=\pi x \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=4$. So the interval of integration is $[0,4]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^4 \left(\pi x\right)dx \\\\ &=\pi \int_0^4x\, dx \end{aligned}$ Let's evaluate the integral. $\pi \int_0^4x\, dx=8\pi$ In conclusion, the volume of the solid is $8\pi$.